Question

A 36.0µC charge is placed at (-0.409m, 0.439m) and a -44.0µC is then placed at (0.154m, 0.423m). What would be the net force on a 25.0µC charge placed at the origin of this coordinate system due to the other two charges?

a) Magnitude _____ N

b) Angle in respect to the positive x-axis _______ degrees

Answer 1

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Solution - -4hMG (0.154 m, 0.423m) 25MC 1(0,0) 36M C (-0.409 m, 0.439 m) distance Now, force between - 44 MC and 25MC . 81= √(0.154-0)²+(0.423-0) ² 0.450m Now distance between 25MC and 36 MC. 8 2 = (-.409 - 0) + (0.439-0)' 0.598 m. Now, - 44MC = -44 X 106c 25 uc 25 x 10-6C 36кс 36 x 10 °C

Now, foru between 25MC and - 4400 F = kg 92 9x10° X 25X10 X 44 X10" - 48.9N (0.450) 22 Now, force between 254C and 36 С. F = kgig2 22.7 N 9x109 x 25x10 6 X 36x10 (0.598) 2 92

Now, angle made by fi with positive x-axis tono i = 0.423 0.154 tan I 0.423 0.154 70° So, angle made by Fi with positive x-axis. 0₂= tani 0.439 -0.409 47.0° So. forces in vector form on 25MC. F = Ficosdi + Firino, i - 48.9 w 70+ 48. 9. sin 70°j 31.0 i + 37.8j Po hodi hindi F2 = F col O (-i) + Fasin Ozj =

= 22.701 47° (1) + 22.7sin 4706) = - 15.5i + 16.6; Now, ruat net force veetor. ol 14 .ר R = Fit F. - 31.00 + 30.0) + (-15.5) - 16.6; 15.5i + 21.2 j (a) Magnitude = 1 R = √ 15.5² +21.2² = 26.3 N Any (6) Angle = tant 21.2 15.5 53.8° Ang